# Open mapping theorem (functional analysis)

Suppose

${displaystyle A:Xto Y}$is a surjective continuous linear operator. In order to prove that

${displaystyle A}$is an open map, it is sufficient to show that

${displaystyle A}$maps the open unit ball in

${displaystyle X}$to a neighborhood of the origin of

${displaystyle Y.}$

Let

${displaystyle U=B_{1}^{X}(0),V=B_{1}^{Y}(0).}$

Then

Since

${displaystyle A}$is surjective:

But

${displaystyle Y}$is Banach so by Baire’s category theorem

That is, we have

${displaystyle cin Y}$and

${displaystyle r>0}$

Let

${displaystyle vin V,}$then

By continuity of addition and linearity, the difference

${displaystyle rv}$satisfies

and by linearity again,

where we have set

${displaystyle L=2k/r.}$

It follows that for all

and all

${displaystyle epsilon >0,}$${displaystyle xin X}$

such that

Our next goal is to show that

${displaystyle Vsubseteq A(2LU).}$

Let

${displaystyle yin V.}$

By (1), there is some

with

${displaystyle left|x_{1}right|$and

${displaystyle left|y-Ax_{1}right|<1/2.}$

Define a sequence

inductively as follows.

Assume:

Then by (1) we can pick

${displaystyle x_{n+1}}$so that:

so (2) is satisfied for

${displaystyle x_{n+1}.}$Let

From the first inequality in (2),

${displaystyle left(s_{n}right)}$is a Cauchy sequence, and since

${displaystyle X}$is complete,

${displaystyle s_{n}}$converges to some

${displaystyle xin X.}$

By (2), the sequence

tends to

${displaystyle y}$and so

${displaystyle Ax=y}$by continuity of

${displaystyle A.}$

Also,

This shows that

${displaystyle y}$belongs to

${displaystyle A(2LU),}$so

${displaystyle Vsubseteq A(2LU)}$ as claimed.

Thus the image

of the unit ball in

${displaystyle X}$contains the open ball

${displaystyle V/2L}$of

${displaystyle Y.}$

Hence,

is a neighborhood of the origin in

${displaystyle Y,}$and this concludes the proof.

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